I have the i.i.d. exponential random variables $X_1, \dots, X_n$ with the density functions
$$f(x; \sigma, \tau)= \begin{cases} \dfrac{1}{\sigma} e^{-(x - \tau)/\sigma} &\text{if}\, x\geq \tau\\ 0 &\text{otherwise} \end{cases}$$
I want to calculate the sufficient statistic for $(\sigma, \tau)$. I calculated the likelihood as follows:
$$\begin{align} L(\sigma, \tau; \mathbf{x}) &= \prod_{i = 1}^n \dfrac{1}{\sigma} \exp{\left\{ \dfrac{-(x_i - \tau)}{\sigma} \right\}} \mathbb{1}_{x \ge \tau} \\ &= \dfrac{1}{\sigma^n} \exp{\left\{ \dfrac{1}{\sigma} \sum_{i = 1}^n -(x_i - \tau) \right\}} \mathbb{1}_{x \ge \tau} \end{align}$$
I would usually then use the Fisher-Neyman factorization theorem and factor this into the form $L(\sigma, \tau; \mathbf{x}) = g(T(\mathbf{x}), (\sigma, \tau)) \times h(\mathbf{x})$, where $T(\mathbf{X})$ is the sufficient statistic. However, the examples that I have seen of people solving this involve some expressions involving lots of $\min$ values. From what I can tell, this occurs because of the $\mathbb{1}_{x \ge \tau}$, which makes sense. However, the problem is that I don't really understand why the various 'steps' are being taken, and so I can't really follow the reasoning. So for such a problem as this, how does one calculate the sufficient statistic for $(\delta, \tau)$? And what is the reasoning involved?
There is an indicator for each observation so it goes like this. The likelihood of a single observation is $$f(x;\sigma,\tau)=\frac 1 \sigma e^{-(x-\tau)/\sigma} \mathbb 1_{x\ge\tau}$$ so the likelihood for our sample is $$\begin{split}L(\sigma,\tau;\textbf x)&=\frac 1 {\sigma^n}\exp\left\{- \frac 1 \sigma\sum_{i=1}^n (x_i-\tau)\right\}\prod_{i=1}^n\mathbb 1_{x_i\ge \tau}\\ &\propto\exp\left\{-\frac 1 \sigma\left(\sum x_i-n\tau\right)\right\}\mathbb 1_{\min(x_i)\ge \tau}\end{split}$$
where the min comes from the fact that the indicators must all be $1$ for the likelihood to be non-zero, otherwise a data point impossible was observed, so the entire likelihood is $0$ (otherwise it just evaluates to the likelihood of the data). So the product expression is the same as we need the smallest observed value to be at least the lower bound.
Thus the jointly sufficient statistic is $\left(\sum_i x_i, \min\{x_i\}\right)$.